A condenser of 2 capacitance is charged steadily from 0 to 5C ....

The charge on a condenser is given by $Q = CV$.

$⇒ V = \dfrac{1}{C}Q$

Since the capacitance depends on the construction of the condenser, it is constant. Therefore, the graph will be straight line with slope $= \dfrac{1}{C}$.
Slope $= \dfrac{1}{C} = \dfrac{1}{2 \times {10}^{- 6}} = 5 \times {10}^{5}$