Answer
The projectile will reach a height of $~~2.5\times 10^7~m~~$ above the surface.
Work Step by Step
We can use conservation of energy to find the distance the projectile reaches from the Earth's center:
$K_f+U_f = K_i+U_i$
$0+U_f = K_i+U_i$
$-\frac{GMm}{r_f} = \frac{1}{2}mv^2-\frac{GMm}{R_E}$
$-\frac{2GM~R_E}{r_f} = v^2~R_E-2GM$
$r_f = -\frac{2GM~R_E}{v^2~R_E-2GM}$
$r_f = \frac{2GM~R_E}{2GM-v^2~R_E}$
$r_f = \frac{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)~(6.37\times 10^6~m)}{(2)(6.67\times 10^{-11}~N~m^2/kg^2)(5.98\times 10^{24}~kg)-(1.0\times 10^4~m/s)^2~(6.37\times 10^6~m)}$
$r_f = 3.16\times 10^7~m$
We can find the height reached above the surface:
$h = (3.16\times 10^7~m)-(6.37\times 10^6~m) = 2.5\times 10^7~m$
The projectile will reach a height of $~~2.5\times 10^7~m~~$ above the surface.
