This is a somewhat lengthy explanation (and using different values) which confirms Keith’s original solution.
You have only been given a fixed numerical value for the TAS, so all other numerical values for the Outbound Track must be based on assumption.
That’s not as ambiguous as it seems at first glance because you are told that the effective Wind Component is +45 kts, making the GS 240 + 45 = 285 kts.
You are also told that the Drift is -15°. Working on the basis that Drift is invariably FROM Heading TO Track, choose some simple numbers to make your calculations easier: I used Heading = 115°T and Track = 100°T.
With these four components we can now establish the Wind Vector.
On the CRP-5 (HI SPEED):
Place centre dot over TAS = 240 kts
Rotate central bezel to place HEADING = 115°T on inner scale under TRUE HEADING index on outer scale
Make a Wind Mark where the straight, vertical Drift line (=15° LEFT) crosses the curved, horizontal speed line at 285 kts
Rotate the central bezel to place the Wind Mark on the central Drift line UNDER the TAS
Read the Wind Direction under the TRUE HEADING index = 230°
Read the Wind Speed by counting down to the Wind Mark from the TAS = 80 kts
Wind Vector = 230° / 80 kts
To calculate the inbound effective wind component, we note that we have a W/V of 230/80, a TAS of 240 kts and a reciprocal Track of 280°T (100°T + 180°)
On the CRP-5 (HI SPEED):
Place centre dot over TAS = 240 kts
Rotate central bezel to place Wind Direction = 230°T on inner scale under TRUE HEADING index on outer scale
Make a Wind Mark 80 kts BELOW the TAS at 160 kts (240 – 80)
Place Track = 280°T on inner scale under TRUE HEADING index on outer scale
Note Wind Mark has moved to indicate 18° of Right Drift
Align 280°T on inner scale with Drift = 18°L on outer scale
Align Track = 280°T on inner scale with Drift = 18°L on outer scale
Note Wind Mark has moved to indicate 14° of Right Drift
Align Track = 280°T on inner scale with Drift = 14°L on outer scale
Note Wind Mark has moved to indicate 15° of Right Drift
Align Track = 280°T on inner scale with Drift = 15°L on outer scale
Note Wind Mark still indicates 15° of Right Drift
The Drift on the outer scale now balances the Drift on the Drift grid, meaning the computation is complete
Read Heading = 265T under the TRUE HEADING Index
Read Ground Speed = 177 kts (ish) on curved speed line
Difference between TAS and GS = 240 kts – 177 kts = 63 kts
GS is slower than TAS, therefore, effective wind component = -63 kts
Closest answer = -65 kts
NOTE that it you are completing these questions under the EASA Learning Objectives, you are required you to use a Navigation Computer and NOT trigonometry. In fact, there ARE some CQB questions where the mathematically more precise trigonometrical solution has been provided as an incorrect option. Bonkers maybe but don't shoot me, I am merely the messenger, I get enough flak for the examiner in class off the students!