= Solution
Using the stipulated point-mass quadrupole, take $z(t)=L\cos\omega t$ and $Q_{ij}=m(x_ix_j-r^2\delta_{ij}/3)$. The only nonzero components are
$$
Q_{xx}=Q_{yy}=-\frac m3z^2,\qquad Q_{zz}=\frac{2m}3z^2.
$$
Since $z^2=L^2(1+\cos2\omega t)/2$, its third <derivative> is $4L^2\omega^3\sin2\omega t$. Therefore
$$
\left\langle\dddot Q_{ij}\dddot Q_{ij}\right\rangle=\frac23m^2\left\langle[(z^2)''']^2\right\rangle=\frac{16}{3}m^2L^4\omega^6,
$$
and the total outward radiated power is
$$
\boxed{\left\langle P_{\rm GW}\right\rangle=\frac{16}{15}m^2L^4\omega^6\quad(G=c=1),\qquad
\frac{16Gm^2L^4\omega^6}{15c^5}\quad\text{in physical units}.}
$$
Quadrupole radiation oscillates at twice the motion frequency. The approximation assumes slow motion $L\omega\ll c$ and a prescribed motion relative to the rest of the source. A lone freely moving point mass cannot execute this oscillation; a physical conserved source includes the accelerating apparatus or other matter, whose quadrupole would also need specification. The displayed value is the point-mass contribution requested by the supplied formula.
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