OK, let's call the transmitter impedance ZT=a+jb where a=150 ohm and b=75 ohm
then let's call the antenna impedance ZA=m+jn where m=75 ohm and n=15 ohm
Furthermore we know that Xc=1/(wC) and XL=wL.
Let's calculate the impedance seen by the antenna. The transmitter has a Thevenin generator in series with the ZT. Short this generator.
We have ZT in parallel with the matching capacitor, all this in series with the inductor, then the impedance seen by the antenna will be:
Zx=(a+jb)*jXc/[a+j(b+Xc)] + jXL
It's quite easy to find the real and inmaginary part of Zx that is:
real(Zx)=a*Xc^2/[a^2+j(b+Xc)^2]
imag(Zx)=[a^2*Xc+b*Xc*(b+Xc)]/[a^2+j(b+Xc)^2]
we can notice that the real part doesn't depend from XL, but only from Xc. So we can solve with respect to Xc:
real(Zx)=m
I obtained Xc=-108.8 ohm, that is C=0.73 pF @ 2GHz
now we have to solve
imag(Zx)=n
after substituting in Zx the value of Xc previously calculated I've found XL=106.9 ohm, that is L=8.5 nH @ 2GHz
This value is different from that calculated in the solution you gave. But if n=-15 ohm, instead of +15 ohm I've found XL=76.9 ohm, that is L=6.12 nH @ 2GHz