The value of $q$ can be calculated using $V = kq/r$. Solving for $q$ and substituting the known values, we find

$$
begin{align*}
q &= frac{rV}{k}\
&= frac{(1.1;mathrm{m})(2.8times10^4;mathrm{V})}{8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C}^2}\
&= 3.4;mumathrm{C}
end{align*}
$$

$3.4;mu$C

Let $r$ be the distance at which the electric potential due to the charge is $1.5times10^4$ V. The value of $r$ can be found using the relationship $V = kq/r$. Solving for $r$ and substituting the known values, we find

$$
begin{align*}
r &= frac{kq}{V}\
&= frac{(8.99times10^{9};mathrm{N}cdotmathrm{m}^2/mathrm{C}^2)(2.9times10^{-6};mathrm{C})}{1.5times10^{4};mathrm{V}}\
&= 1.7;mathrm{m}
end{align*}
$$

Applying $PE = kqq_0/r$ with $q = 6.8;mu$C, $q_0 = 4.4;mu$C and $r = 0.13$ m gives

$$
begin{align*}
PE &= (8.99times10^9;mathrm{N}cdotmathrm{m}^2/mathrm{C}^2)timesfrac{(6.8times10^{-6};mathrm{C})(4.4times10^{-6};mathrm{C})}{0.13;mathrm{m}}\
&= 2.1;mathrm{J}
end{align*}
$$