= Solution
Interpret the stated side lengths as all eight edges of the <square pyramid>. The equal lateral edges put its apex above the base <centre>. The base-centre-to-vertex distance is half the diagonal, $1/\sqrt2\,\mathrm m$. The <Pythagorean theorem> therefore gives
$$
h=\sqrt{1^2-(1/\sqrt2)^2}=\frac1{\sqrt2}\,\mathrm m.
$$
Using the <pyramid> volume formula with a $1\,\mathrm{m^2}$ base,
$$
\boxed{V=\frac13(1)\frac1{\sqrt2}
=\frac1{3\sqrt2}\,\mathrm{m^3}\approx0.2357\,\mathrm{m^3}.}
$$
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