= Solution
Let $\widetilde A=\widehat U\widehat\Sigma\widehat V^T$. Interpret the diagonal of $\widehat\Sigma$ as nonnegative computed <singular values> and sort them consistently. If signed diagonal entries are used instead, take their absolute values and absorb signs into the <orthonormal> columns. Because the two outer factors have the stated <orthonormality>, these sorted numbers are exactly $\sigma_j(\widetilde A)$.
For completeness, the <singular value perturbation bound> follows from the <Eckart-Young-Mirsky theorem>. If $B_{j-1}$ is the truncated <singular value decomposition> of a <matrix> $B$, the <triangle inequality> gives
$$
\sigma_j(A)=\min_{\operatorname{rank}D\leq j-1}\|A-D\|_2
\leq\|A-B_{j-1}\|_2
\leq\|A-B\|_2+\sigma_j(B).
$$
Exchange $A$ and $B$ to get $|\sigma_j(A)-\sigma_j(B)|\leq\|A-B\|_2$. Applying this to $B=\widetilde A$ and the given <backward error> proves
$$
\boxed{|\sigma_j(A)-\widehat\sigma_j|\leq\epsilon\quad(1\leq j\leq n)}.
$$
Thus <backward stability> of the <singular value decomposition> gives an absolute <forward error> bound for every <singular value>, even a very small one.
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