Assuming 2032 is itself a number in base 4, you have
\(2032_4=2\times4^3+3\times4^1+2\times4^0\)
Or, if you mean to ask about converting 2032 into base 4, you have
\(\dfrac{2032}4=508\) with remainder 0, which means the "ones" digit is 0.
\(\dfrac{508}4=127\) with remainder 0, which means the "tens" digit is also 0.
\(\dfrac{127}4=31\) with remainder 3, so the "hundreds" digit is 3.
\(\dfrac{31}4=7\) with remainder 3, so the "thousands" digit is also 3.
\(\dfrac74=1\) with remainder 3, so the next digit is also 3.
\(\dfrac14=0\) with remainder 1, so the next (and final) digit is 1.
So, \(2032_{10}=133300_4\).
