= Solution
The <band effective mass> is determined by the <second derivative> of the <dispersion relation>:
$$
\frac1{m^*(k)}=\frac1{\hbar^2}\frac{d^2E}{dk^2}.
$$
With the stated sign convention and $t>0$, the band minimum is at $k=0$. Expanding the <cosine>, or differentiating twice, gives
$$
E(k)=E_0-2t+ta^2k^2+O(k^4),\qquad
\left.\frac{d^2E}{dk^2}\right|_0=2ta^2.
$$
Comparison with a free <Electron>'s <kinetic energy> $\hbar^2k^2/(2m_e)$ therefore requires
$$
\boxed{t=\frac{\hbar^2}{2m_ea^2}
=6.78\times10^{-20}\,\mathrm J
=0.423\,\mathrm{eV}.}
$$
The <hopping integral> has units of <energy>. A negative $t$ would move the minimum to the edge of the <Brillouin zone> and give the same positive bottom-of-band <effective mass> with $|t|$ in this expression.
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