= Solution
Write every displayed block identity as $I_2$. Direct block multiplication in the <Dirac representation of the gamma matrices> gives
$$
(\widetilde\gamma^0)^2=I_4,\qquad
\widetilde\gamma^0\widetilde\gamma^i=\begin{pmatrix}0&\sigma^i\\\sigma^i&0\end{pmatrix},\qquad
\widetilde\gamma^i\widetilde\gamma^0=\begin{pmatrix}0&-\sigma^i\\-\sigma^i&0\end{pmatrix}.
$$
Thus their mixed <anticommutator> is zero. For two spatial indices,
$$
\widetilde\gamma^i\widetilde\gamma^j=-\begin{pmatrix}\sigma^i\sigma^j&0\\0&\sigma^i\sigma^j\end{pmatrix}.
$$
Using the <Pauli matrix multiplication law> gives
$$
\boxed{\{\widetilde\gamma^0,\widetilde\gamma^0\}=2I_4,\quad\{\widetilde\gamma^0,\widetilde\gamma^i\}=0,\quad\{\widetilde\gamma^i,\widetilde\gamma^j\}=-2\delta^{ij}I_4.}
$$
These exhaust all time-time, time-space and space-space cases of the <Clifford algebra>. No special choice of individual <Pauli matrices> beyond their stated <anticommutator> is needed.
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