$48^{\circ}$

As $PQRST$ is a regular pentagon, each of its internal angles is $108^{\circ}$. The internal angles of the quadrilateral $PRST$ add up to $360^{\circ}$ and so by symmetry $\angle PRS=\angle RPT=\frac{1}{2}(360^{\circ}-2\times 108 ^{\circ})=72^{\circ}$. Each interior angle of a regular hexagon is $120^{\circ}$, so $\angle PRU=120^{\circ}$.\par Therefore $\angle SRU=\angle PRU-\angle PRS=120^{\circ}-72^{\circ}=48^{\circ}$.

*This problem is taken from the UKMT Mathematical Challenges.*

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