According to the ones digit of perfect squares, we can eliminate choice $B$.  A perfect square ending with $5$ must be a multiple of $25$, so we can eliminate choice $A$.  $${{48}^{2}}=2304$$ and $${{49}^{2}}=2401$$ so we can eliminate choice $C$.  ${57}^{2}=3249$. We choose option $D$. 
$$\overline{**45}$$, $$\overline{19*8}$$, $$\overline{23*1}$$, and $$\overline{3*49}$$ are four $4-$digit numbers with some unknown digits. Which number is possible to be a perfect square? 
Options:
[a] $$\overline{**45}$$ 
[b] $$\overline{19*8}$$ 
[c] $$\overline{23*1}$$ 
[d] $$\overline{3*49}$$ 
[e] None 