Given,
Resistance of the element, $\mathrm{R} = 117 \Omega$  
Resistance of heating element at room temperature, ${\mathrm{R}}_T = 100 \Omega$
Temperature coeffient of the material of resistor, $\mathrm{\alpha }$= 1.70 x 10^–4 °C^–1

Now, using the relation

                   $\boxed{\mathrm{R} = {\mathrm{R}}_{\mathrm{T} }\left[1+\mathrm{\alpha } (\mathrm{t}-27)\right]}$
we get,
                $$\begin{gathered} 117 = 100 [1+1.70 \times {10}^{-4} (\mathrm{t}-27\left)\right] \\ \mathrm{t} = 1000+27 = 1027°\mathrm{C} \end{gathered}$$ 
is the required temperature of the element.

Given,
EMF of battery, $\mathrm{\epsilon } = 10 \mathrm{V}$ 
Internal resistance of battery, r= $3 \mathrm{\Omega }$ 
Current flowing in the circuit, I= 0.5 A 

Using the forlmula   $\boxed{\mathrm{I} = \frac{\mathrm{\epsilon }}{\mathrm{R}+\mathrm{r}}}$ 

or                           $\mathrm{R} = \frac{\mathrm{\epsilon }}{\mathrm{I}}-\mathrm{r}$ 
where, R is the external resistance.
$\mathbf{\therefore }$       $\mathrm{R} = \frac{10}{0.5}-3 = 17 \mathrm{\Omega }$ is the required resistance.

Terminal voltage,  $$\begin{gathered} \mathrm{V} = \mathrm{IR} \\ = 0.5 \times 17 \\ = 8.5 \mathrm{V} \end{gathered}$$