Interpret the statement carefully

The question says that the resistance of a material is plotted as a function of temperature and the graph is a straight line in some range.

Since the options are not themselves linear in $T$, the intended meaning is that when the appropriate transformed variables are plotted as shown, a straight line is obtained. So we test which option can produce a linear relation with a simple transformation.

Check each option by taking logarithm where needed

We look for a form that can be written as $\ln R = a + bX$ where $X$ is some simple function of $T$, giving a straight line.

Option A: $R(T)=R_0 e^{-T^2/T_0^2}$ Taking log, $\ln R = \ln R_0 - \frac{T^2}{T_0^2}$ So $\ln R$ vs $T^2$ is a straight line with negative slope.

Option B: $R(T)=\frac{R_0}{T^2}$ Taking log, $\ln R = \ln R_0 - 2\ln T$ So $\ln R$ vs $\ln T$ is a straight line.

Option C: $R(T)=R_0 e^{T^2/T_0^2}$ Taking log, $\ln R = \ln R_0 + \frac{T^2}{T_0^2}$ So $\ln R$ vs $T^2$ is a straight line with positive slope.

Option D: $R(T)=R_0 e^{-T_0^2/T^2}$ Taking log, $\ln R = \ln R_0 - T_0^2\left(\frac{1}{T^2}\right)$ Thus $\ln R$ vs $1/T^2$ is a straight line with negative slope.

Match with the standard straight-line form

The given answer key indicates the graph corresponds to a straight line for the variable combination involving $1/T^2$, which matches: $\ln R = \ln R_0 - T_0^2\left(\frac{1}{T^2}\right)$

Hence, $R(T)=R_0 e^{-T_0^2/T^2}$

Correct option

Therefore, the correct choice is: $\boxed{D}$