jmc

If we expand this using the binomial theorem, we get a bunch of terms with $\sqrt{3}$ in them. To avoid painful estimation, we use the following trick: Add $(2-\sqrt{3}{)}^4$ to this expression. We know that $(2-\sqrt{3}{)}^4$ is small, since $2-\sqrt{3}<1$. When we add these together, the $\sqrt{3}$ terms magically cancel out. By the Binomial Theorem, $$(2+\sqrt{3}{)}^4=2^4+4\cdot 2^3\cdot (\sqrt{3})+6\cdot 2^2\cdot (\sqrt{3}{)}^2+4\cdot 2\cdot (\sqrt{3}{)}^3+(\sqrt{3}{)}^4$$ whereas $$(2-\sqrt{3}{)}^4=2^4-4\cdot 2^3\cdot (\sqrt{3})+6\cdot 2^2\cdot (\sqrt{3}{)}^2-4\cdot 2\cdot (\sqrt{3}{)}^3+(\sqrt{3}{)}^4$$ Therefore, their sum is $$2(2^4+6\cdot 2^2(\sqrt{3}{)}^2+(\sqrt{3}{)}^4)=2(16+72+9)=194$$ Since the term we added, $(2-\sqrt{3}{)}^4$, is less than a half (actually, it's less than .01), $\boxed{194}$ is the closest integer to $(2+\sqrt{3}{)}^4$.