According to the question the final image is formed at infinity.
The object for eyepiece (image formed by objective lens) should be at a distance of f = 3 cm from it.
Hence distance of the image formed by the objective lens from the objective lens is 15 − 3 = 12 cm
Using lens formula for the objective lens,
\(\frac 1f = \frac 1v - \frac 1u\)
\(\frac 12 = \frac 1{12} - \frac 1x\)
x = −2.4 cm
-ve indicates distance is towards left of the lens.