Shift, invert, cube

Idea. Shifting, inverting and cubing a matrix do the same to each eigenvalue, with the same eigenvectors. So the eigenvalues of $B$ are $(\lambda -3{)}^{-3}$, and the trace is their sum.

Step 1: eigenvalues of $B$. If $Av=\lambda v$, then $(A-3I)v=(\lambda -3)v$, so $(A-3I{)}^{-1}v=\frac{1}{\lambda -3}v$ and $Bv=\frac{1}{(\lambda -3{)}^3}v$. The shift keeps the matrix invertible, since $5-3$ and $7-3$ are not 0.

Step 2. The eigenvalues of $B$ are $\frac{1}{2^3}=\frac{1}{8}$ and $\frac{1}{4^3}=\frac{1}{64}$.

Step 3. $$\mathrm{tr}B=\frac{1}{8}+\frac{1}{64}=\frac{9}{64}.$$

Remark. The argument applies even if $A$ is not diagonalisable: for any polynomial or rational function $f$ defined at the eigenvalues, the eigenvalues of $f(A)$ are $f({\lambda }_i)$, counted with multiplicity.