= Solution
Write $\int_{\mathbf p}=\int d^3p/(2\pi)^3$ and $E_{\mathbf p}=\sqrt{\mathbf p^2+m^2}$. Insert the <mode expansion of a free field> for each quantity into the <commutator>. Only the mixed <creation operator> and <annihilation operator> terms survive. The <Dirac delta function> in their <canonical commutation relation> sets the two <momenta> equal, so the square-root factors multiply to $1/2$:
$$
[\phi(\mathbf x),\pi(\mathbf y)]
=\frac i2\int_{\mathbf p}\left[e^{i\mathbf p\cdot(\mathbf x-\mathbf y)}+e^{-i\mathbf p\cdot(\mathbf x-\mathbf y)}\right]
=\boxed{i\delta^3(\mathbf x-\mathbf y)}.
$$
In the second equality, each <Fourier transform> is the same <Dirac delta function>, since the integration domain is invariant under $\mathbf p\mapsto-\mathbf p$. This is the equal-time <canonical commutation relation>. The analogous calculation gives $[\phi(\mathbf x),\phi(\mathbf y)]=[\pi(\mathbf x),\pi(\mathbf y)]=0$, because their two mixed terms have opposite signs.
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