To understand $u_1$, we must first find out what the register $r_5$ is storing.
If you check closely the line $L_{02}$, it’s fetching an element from array $b$ and storing it in $r_5$.
Line $L_{04}$ is storing the value present in $r_5$ to array $a$, so we’re definitely performing the multiplication by $8$ in $L_{03}$.
$L_{03}$ is a shift-left operator. To multiply by $8$, how many bits (in binary) do we need to shift to left? If you can find out that number, that would be the answer to $u_1$.
$L_{05}$ and $L_{06}$ are incrementing the values of registers $r_3$ and $r_5$ respectively to point to next element of $a$ and $b$ respectively. But by how much shall we increase it to make it point to next element?
Suppose at some point in time, $r_3$ is storing $1000$ in decimal and pointing to 2nd element of array $a$. Recall that each element is 4Bytes in size (given in question). Now, what is the address of 3rd element of array $a$, can you guess? Yes, it would be $1004$. Hence, $4$ must be added.
Note: Given memory is byte addressable. Each int takes $32 bit = 4B$, so we need to shift by 4 addresses.
$u_4$ should be the line to make the loop repeat. Note that if we make it $L_{02}$, the loop termination condition is skipped and would lead to infinite loop (or unless we got trapped due to accessing illegal memory).
The value of $u_1, u_2, u_3, u_4$ will be $3, 4, 4, L01$.
Option B is the right answer.
