As the data is not complete, a similar question is found online and the missing data is used from that question. Similar question is attached herewith.

From the given data

The allowed total strain is given as $\delta_a=0.0371 mm$

The length of the rod is missing in this data which in the additional data is l=290 mm

The Modulus of Elasticity is given as E=210 GPa

The coefficient of thermal expansion is given as α= 10 ×10⁻⁶ mm / mm⋅°C.

The initial temperature is room temperature which is given as T1=22C

The final temperature is given as T2=140 C

The strain due to stress is given as

$\delta_s=\dfrac{\sigma l}{E}$

$\delta_s=\dfrac{\sigma (0.290)}{210\times 10^9}\\\delta_s=1.38095\sigma \times 10^{-12}$

Now the thermal strain is given as

$\delta_c=l\alpha \Delta T\\\delta_c=0.290\times 10 \times 10^{-6}(T_2-T_1)\\\delta_c=0.290\times 10 \times 10^{-6}(140-22)\\\delta_c=0.0003422$

So now the equation of the allowed strain is given as

$\delta_a=\delta_s+\delta_c\\0.0371\times 10^{-3}=1.38095\sigma \times 10^{-12}+0.0003422\\\dfrac{1.38095}{10^{12}}\sigma+0.0003422=\frac{0.0371}{1000}\\\sigma=-\dfrac{10^{12}\times \:0.0003051}{1.38095}\\\sigma=-220934863.68 Pa\\\sigma=-220.9 MPa$