Answer
$\rm 16.6\;km/s$
Work Step by Step
Let's assume that the projectile is a rocket and we will launch it in the same direction as Earth's speed around the sun.
So first, we need to find the rocket's escape velocity from the sun when it is at the same distance from it as the Earth.
We assume that the sun-rocket system is isolated and no external forces are exerted on our system [we neglected the Earth's gravitational pull here].
$$E_i=E_f$$
$$K_i+U_{ig}=K_f+U_{fg}$$
The final gravitational potential energy of the rocket when it at an infinite distance from the sun is zero and so does its final kinetic energy since it finally stops.
$$K_i+U_{ig}=0+0$$
$$\frac{1}{2}\color{red}{\bf\not}m_Rv_{\rm escape }^2+\dfrac{-GM_{\rm Sun}\color{red}{\bf\not}m_{R}}{R_{\rm E\rightarrow S}}=0+0$$
where $m_R$ is the rocket's mass and $R_{\rm E\rightarrow S}$ is the distance between the Earth and the Sun.
Hence, the escape velocity from the sun is given by
$$v_{\rm escape }=\sqrt{\dfrac{2GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}\tag 1$$
Now we know that the Earth is moving with some velocity around the sun. This Earth's speed is the initial speed of the rocket before launching it away from the Earth and then away from the Solar system.
The rocket's escape velocity $v_{\rm escape }$ from the solar system is the sum of the Earth's speed plus its speed after escaping from Earth's pull.
$$v_{\rm escape }=v_E+v_{R}$$
Thus,
$$v_R=v_{\rm escape }-v_E\tag 2$$
where $v_E$ is Earth's speed around the sun and $v_{R}$ is the rocket's speed away from Earth after escaping from its gravitational pull.
Now we need to find $v_E$ by using Newton's second law:
$$\sum F_r=F_{G,E\rightarrow S}=m_Ea_r$$
$$\dfrac{GM_{\rm Sun}\color{red}{\bf\not}m_E}{R_{\rm E\rightarrow S}^2}=\dfrac{\color{red}{\bf\not}m_Ev_E^2}{\color{red}{\bf\not}R_{\rm E\rightarrow S}}$$
$$v_E=\sqrt{\dfrac{GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}\tag 3$$
Noting that we ignored the rotating speed of Earth around its own axis here as the author told us.
Plugging (1) and (3) into (2);
$$v_R=\sqrt{\dfrac{2GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}-\sqrt{\dfrac{GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}\tag 4$$
Now we need to find the rocket's launching speed from the earth. We will use the conservation of energy as well here as the rocket and the earth are our new system.
$$E_i=E_f$$
$$K_i+U_{ig}=K_f+U_{fg}$$
The rochet's final gravitational potential energy relative to the earth is zero.
$$K_i+U_{ig}=K_f+0$$
$$\frac{1}{2}\color{red}{\bf\not}m_Rv_{\rm Launch}^2+\dfrac{-Gm_E\color{red}{\bf\not}m_R}{R_E}=\frac{1}{2}\color{red}{\bf\not}m_Rv_R^2$$
where $R_E$ is the Earth's radius since it will be launched from the surface of the earth.
Hence, the Launching speed is given by
$$v_{\rm Launch}=\sqrt{v_R^2+\dfrac{2Gm_E}{R_E}}$$
Plugging $v_R$ from (4);
$$v_{\rm Launch}=\sqrt{v_R^2+\dfrac{2Gm_E}{R_E}}$$
Plugging from (4);
$$v_{\rm Launch}=\sqrt{\left(\sqrt{\dfrac{2GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}-\sqrt{\dfrac{GM_{\rm Sun}}{R_{\rm E\rightarrow S}}}\right)^2+\dfrac{2Gm_E}{R_E}}$$
Plugging the known:
$$v_{\rm Launch}=\sqrt{\left(\sqrt{\dfrac{2(6.67\times 10^{-11})(1.99\times 10^{30})}{(1.5\times 10^{11})}}-\sqrt{\dfrac{(6.67\times 10^{-11})(1.99\times 10^{30})}{(1.5\times 10^{11})}}\right)^2+\dfrac{2(6.67\times 10^{-11})(5.98\times 10^{24})}{ 6.37\times 10^6}}$$
$$v_{\rm Launch}=\color{red}{\bf 16.6\times10^3}\;\rm m/s$$
