$\underline{\mathbf{Answer:}}\;\bbox[lightgreen, 5px, border: 2px solid black]{\color {black} {160 \times 10^6\; \frac{\mathrm {Bytes}}{\mathrm{sec}}}}$

$\underline{\mathbf{Explanation:}}$

$\text {Given frequency} = 60\ \text{MHz}$

This means that the processor completes $\color{green}{60\times 10^6 \;\text{cycles in}\;1 \;\text{second}}.$

$\therefore $ One cycle is completed in $\dfrac{1}{60\times 10^6 }\; \text{seconds}$

Now,

To service a cache miss, number of cycles needed $ = 1\;\text{cylce}\;\color{blue}{\text{(to accept starting address of the block)}} + 3\;\text{cylces}\;\color{blue}{\text{(to fetch all the $8$ words of the blocks)}} + \underbrace{8\times1}_\text{$\because$1 word per cycle}\;\text{cylces}\;\color{blue}{\text{(to transmit all $8$ words of the block)}}= 12 \;\text{cycles}$

$\mathbf{Note:}$ Total data is the data which is used for trasmitting the words of the requested block at the rate of $\mathbf{1}$ word per cycle $=\mathbf{8\;words \times 4\;Byte\;(Size\; of\; each\; word)}$

$ \begin{align}\therefore \mathbf{Bandwidth} \require{cancel} &= \dfrac{\text{Total Data}}{12 \;\text{cycle time}} \\&= \dfrac{8\times4\;\text{Bytes}}{12\;\text{cycles }\times1\; \text{cycle time}} \\&= \dfrac{32}{12 \times \dfrac{1}{60\times10^6}} \\&= \require{cancel} \dfrac{32\times \cancel {60}^{5} \times 10^6}{\cancel{12}^1}\\& = 5 \times 32 \times {10}^6 \\&= \color {black}{160 \times 10^6 \; \dfrac{\text{Bytes}}{\mathrm{sec}}}\end{align}$

$\therefore \bbox[lightgreen, 5px, border: 2px solid black]{\color {black}{160 \times 10^6 \; \dfrac{\text{Bytes}}{\mathrm{sec}}}}$ is the correct answer.