Electric field due to a charged rod along its axis is given by
[tex]E = \frac{kQ}{(L+r)(r)}[/tex]
here we know that
L = 14 cm
r = distance from end of rod
r = 36 - 7 = 29 cm
Q = 222 mC
now we will have
[tex]E = \frac{(9 \times 10^9)(222 \times 10^{-3})}{(0.29)(0.43)}[/tex]
[tex]E = 1.6 \times 10^{10} N/C[/tex]