The expression for current in a p-n junction diode, is given as
$I=I_{0} \exp \left(\frac{e V}{2 k_{B} T}-1\right)$
Here, $l_{0}=5 \times 10^{-12} \mathrm{~A}$
$\mathrm{T}=300 \mathrm{~K}$
$\mathrm{k}_{\mathrm{B}}$ is the Boltzmann constant with a value of $8.6 \times 10^{-5} \mathrm{eV} / \mathrm{k}=8.6 \times 10^{-5} \times 1.6 \times 10^{-19}=1.376 \times 10^{-23} \mathrm{~J} / \mathrm{K}$
(a) Forward voltage is given as $V=0.6 \mathrm{~V}$
$I=5 \times 10^{-12} \exp \left(\frac{1.6 \times 10^{-19} \times 0.6}{2 \times 1.376 \times 10^{-23} \times 300}-1\right)$
$I=5 \times 10^{-12} \exp (22.36)$
$=0.0256 \mathrm{~A}$
(b) Voltage across the diode is given to get increased to $0.7 \mathrm{~V}$
$I^{\prime}=5 \times 10^{-12} \exp \left(\frac{1.6 \times 10^{-19} \times 0.7}{2 \times 1.376 \times 10^{-25} \times 300}-1\right)$
$I^{\prime}=5 \times 10^{-12} \exp (26.25)$
$l^{\prime}=1.257 \mathrm{~A}$
Change in current will be $\Delta \mathrm{l}=\mathrm{l}^{\prime}-\mathrm{I}$
$=1.257-0.0256$
$=1.23 \mathrm{~A}$
Change is voltage will be $0.7-0.6=0.1 \mathrm{~V}$