Numerical experiments suggest that $$\sum_{k=1}^{n-1} k^2\sigma(k)\sigma(n-k) = \frac{n^2}{8}\sigma_3(n) - \frac{4n^3-n^2}{24}\sigma(n).$$ PS. In fact, it directly follows from the quoted Touchard and Ramanujan identities.
A couple of similar identities: $$\sum_{k=1}^{n-1} k\sigma(k)\sigma(n-k) = \frac{5n}{24}\sigma_3(n) - \frac{6n^2-n}{24}\sigma(n).$$ $$\sum_{k=1}^{n-1} k^3\sigma(k)\sigma(n-k) = \frac{n^3}{12}\sigma_3(n) - \frac{3n^4-n^3}{24}\sigma(n).$$