An engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{5}$. When $\mathrm{T}_2$ is lowered by 45 K , its efficiency becomes $\frac{1}{2}$. Temperatures $\mathrm{T}_1$ and $\mathrm{T}_2$ are respectively

The Carnot efficiency formula gives $\eta = 1 - T_2/T_1$ where $T_1$ is the source temperature and $T_2$ is the sink temperature.

For initial efficiency $\eta = 1/5$, solving $1/5 = 1 - T_2/T_1$ yields $T_2/T_1 = 4/5$, so $T_1 = (5/4)T_2$.

When $T_2$ decreases by 45 K with constant $T_1$, the new efficiency is $1/2 = 1 - (T_2 - 45)/T_1$.
This gives $(T_2 - 45)/T_1 = 1/2$, so $T_1 = 2(T_2 - 45)$.

Equating both expressions: $(5/4)T_2 = 2(T_2 - 45)$. Multiplying through by 4: $5T_2 = 8T_2 - 360$.
Solving yields $-3T_2 = -360$, so $T_2 = 120$ K.

Substituting back: $T_1 = (5/4)(120) = 150$ K.

The temperatures are $T_1 = 150$ K and $T_2 = 120$ K, corresponding to option $\boxed{\text{D}}$.