Answer is C.
For $1$ sec it is $10^9$ bytes ( Note - by looking at the options, it can be decided that $1GB$ should be considered as $10^9$, but not as $2^{30}$, and in general, if bandwidth is given, then we consider $1\text{ Giga} = 10^9)$
So, for $64$ bytes?
It is $\dfrac{64\times 1 }{10^9}$ so it is $64\text{ ns}$ but mm latency is $32.$
So, total time required to place cache line is $64+32 = 96\text{ ns}.$