Let's calculate the number of cycles each instruction takes before reaching the $\text{'}$sw$\text{'}$ instruction for the first loop iteration and find the corresponding PC value.

Here's the instruction breakdown with their cycle counts:

$1.$ add TO$,$ zero, zero $=>4$ cycles

$2.$ add T$1,$ R$1,$ zero $=>4$ cycles

$3.$ add T$2,$ R$2,$ zero $=>4$ cycles

$4.$ Loop: lw T$4,0($T$2)=>5$ cycles

$5.$ add T$5,$ T$4,$ C $=>4$ cycles

$6.$ sw T$5,0($T$1)=>4$ cycles $($the one we need to calculate the PC for$)$

Now, we add up the cycles before the $\text{'}$sw$\text{'}$ operation:

$4($add TO$)+4($add T$1)+4($add T$2)+5($lw T$4)+4($add T$5)=21$ cycles

If we're given PC value increments of $4,$ which is typical in MIPS $($as most instructions are $4$ bytes$),$ and we know the loop starts at an initial PC value of $120,$ we can calculate the PC value for the $\text{'}$sw$\text{'}$ instruction as follows:

Starting PC $=120$

PC after $1$st instruction $($add TO$)=120+4=124$

PC after $2$nd instruction $($add T$1)=124+4=128$

PC after $3$rd instruction $($add T$2)=128+4=132$

PC for lw instruction $(4$th$)=132+4=136$

PC for the $\text{'}$sw$\text{'}$ instruction $(6$th$)$ is the one after the 'add T$5\text{'}$ so:

PC for 'add T$5\text{'}(5$th$)=136+4=140$

PC for $\text{'}$sw$\text{'}$ T$5(6$th$)=140+4=144$

Therefore, the correct answer is:

PC value for the SW instruction $=144.$