Answer
$2.77\;\rm km/s$
Work Step by Step
Let's assume that the projectile will travel away from the Earth-moon system to infinity at which its final speed is zero and its final gravitational potential energy is also zero.
We assume that the system is isolated which means that the energy is conserved.
$$E_i=E_f$$
where $E_i$ is the projectile's initial energy when it is escaping the gravity of the Earth-moon system.
$$K_i+U_{pE,i}+U_{pM,i}=K_f+U_{pE,f}+U_{pM,f}$$
$E\rightarrow$ Earth, $M\rightarrow$ Moon, $P\rightarrow$ projectile.
$$\frac{1}{2}m_pv_{escape}^2+\dfrac{-Gm_Em_p}{r_{Ep}}+\dfrac{-Gm_Mm_p}{r_{Mp}}=0+0+0$$
$$ \color{red}{\bf\not}m_pv_{escape}^2+\dfrac{-2Gm_E \color{red}{\bf\not}m_p}{r_{Ep}}+\dfrac{-2Gm_M \color{red}{\bf\not}m_p}{r_{Mp}}=0$$
Hence,
$$v_{escape}^2=\dfrac{ 2Gm_M }{r_{Mp}}+\dfrac{ 2Gm_E }{r_{Ep}}\tag 1$$
From the figure below, we can see that
$$r_{Ep}=R_E+r_{EM}+2R_M$$
And that
$$r_{Mp}=R_M$$
where $R_E$ is Earth's radius, $r_{EM}$ is the distance between the Earth and the moon, and $R_M$ is the moon's radius.
Plugging these two formulas into (1);
$$v_{escape}^2=\dfrac{ 2Gm_M }{R_M}+\dfrac{ 2Gm_E }{R_E+r_{EM}+2R_M} $$
$$v_{escape} =\sqrt {2G\left[\dfrac{ m_M }{R_M}+\dfrac{ m_E }{R_E+r_{EM}+2R_M}\right]} $$
Plugging the known from table 13.2;
$$v_{escape} =\sqrt {2(6.67\times 10^{-11})\left[\dfrac{ (7.36\times 10^{22})}{(1.74\times 10^6)}+\dfrac{ (5.98\times 10^{24})}{(6.37\times 10^6)+(3.84\times 10^8)+2(1.74\times 10^6)}\right]} $$
$$v_{escape} =2.77\times 10^3\;\rm m/s=\color{red}{\bf 2.77}\;\rm km/s$$
