Answer
(a) see prove below.
(b) 106 dB
Work Step by Step
(a) Given $I=\frac{k}{d^2}$ and $B=10log\frac{I}{I_0}$, we have
$B_1=10log\frac{I_1}{I_0}=10log\frac{k/d_1^2}{I_0}=10log\frac{k}{I_0d_1^2}
=10(logk-logI_0-logd_1^2)=10(logk-logI_0)-20logd_1$
which in turn gives $10(logk-logI_0)=B_1+20logd_1$
Similarly, $B_2=10(logk-logI_0-logd_2^2)=10(logk-logI_0)-20logd_2
=B_1+20logd_1-20logd_2$
Thus, $B_2=B_1+20log\frac{d_1}{d_2}$ which proves the relationship.
(b) Given $B_1=120,d_1=2,d_2=10$, use the above relationship, we have
$B_2=120+20log\frac{2}{10}=106$ dB