$\left(132\right)_{4}$
= $(1*4^2 + 3*4^1 + 2*4^0)_{10}$
= $\left(30\right)_{10}$
= $\left(25 + 5 + 0\right)_{10}$ [powers of 5]
= $(1*5^2 + 1*5^1 + 0*5^0)_{10}$
= $\left(110\right)_{5}$
Answer is 110.
Firstly convert the base $4$ into decimal number system (Base $10$):
$(132)_4=(z)_{10} \implies 2*4^0+3*4^1+1*4^2=(30)_{10}$
Now convert $(30)_{10}$ into base $5$,dividing by $5$ we get:
$(30)_{10}=(x)_5$
$\therefore (x)_5=(110)_5$
