= Solution
Use the critical ten-dimensional <RNS string> and <light-cone gauge in string theory>, leaving eight transverse <bosons> and eight transverse <fermions>. The bosonic oscillators $\alpha_{-m}^i$ have <integer> $m>0$. In the <Neveu–Schwarz sector> the fermionic oscillators $b_{-r}^i$ have positive half-integer $r$, while in the <Ramond sector> the fermionic oscillators $d_{-m}^i$ have <integer> $m$, including <zero modes>. Their <anticommutators> are $\{b_r^i,b_s^j\}=\delta^{ij}\delta_{r+s,0}$ and $\{d_m^i,d_n^j\}=\delta^{ij}\delta_{m+n,0}$. The level and mass formulas are
$$
\begin{aligned}
N_{\mathrm{NS}}&=\sum_{m>0}\alpha_{-m}^i\alpha_m^i+\sum_{r>0}r\,b_{-r}^ib_r^i,&\alpha'M^2&=N_{\mathrm{NS}}-\frac12,\\
N_R&=\sum_{m>0}(\alpha_{-m}^i\alpha_m^i+m\,d_{-m}^id_m^i),&\alpha'M^2&=N_R.
\end{aligned}
$$
Before projection the NS <ground state> is a <tachyon> of mass squared $-1/(2\alpha')$. Its oscillator tower describes <spacetime> <bosons>. The <Ramond zero-mode Clifford algebra> $\{\sqrt2d_0^i,\sqrt2d_0^j\}=2\delta^{ij}$ makes the Ramond vacuum a 16-dimensional spinor of $SO(8)$, split into the two chiral eight-dimensional spinors. The Ramond tower describes <spacetime> <fermions>; there is no Ramond <tachyon>.
Choose the supersymmetric <GSO projection>. In the <NS sector> it keeps states with an <odd number> of <fermionic creation operators>, eliminating the zero-oscillator <tachyon>. In the <Ramond sector> it fixes the combined parity of the zero-mode <chirality> and the nonzero-mode <fermion> number. Thus the Ramond vacuum retains one <chirality>; adding an <odd number> of nonzero-mode <fermions> requires the opposite zero-mode <chirality>. The projected towers both have $\alpha'M^2=0,1,2,\ldots$. Here the counts refer to oscillator polarizations for a single endpoint multiplicity; any common Chan-Paton multiplicity multiplies both counts.
At the first retained mass level, $M^2=0$, the NS states are $b_{-1/2}^i|0;k\rangle$ for $i=1,\ldots,8$, furnishing the eight polarizations of the ten-dimensional gauge vector. The Ramond <ground state> is one chiral spinor with eight physical polarizations. Hence
$$
\boxed{M^2=0:\qquad 8\ \text{bosons}=8\ \text{fermions}.}
$$
These form the massless <supersymmetric vector multiplet>.
At the next mass level, $\alpha'M^2=1$, the NS level is $3/2$. The odd-fermion condition leaves exactly
$$
b_{-3/2}^i|0\rangle,\qquad \alpha_{-1}^ib_{-1/2}^j|0\rangle,\qquad b_{-1/2}^ib_{-1/2}^jb_{-1/2}^k|0\rangle\quad(i<j<k).
$$
Their counts are $8$, $8\cdot8=64$ and $\binom83=56$. Anticommutation requires distinct indices in the third family. There are no other partitions of $3/2$ with odd <fermion> number, so these are all 128 <bosons>. They assemble into the massive little-group $SO(9)$ representations of dimensions 44 and 84: a traceless symmetric two-tensor and a three-form.
At Ramond level one, the only possibilities are $\alpha_{-1}^i|s\rangle$ and $d_{-1}^i|s'\rangle$. The spinors $|s\rangle$ and $|s'\rangle$ each have eight components and have opposite zero-mode <chirality> because the fermionic oscillator reverses the GSO parity. Consequently there are $64+64=128$ <fermions>. They form a gamma-traceless vector-spinor of the massive <little group>: a vector-spinor has $9\cdot16$ components and its <gamma trace> removes 16. Therefore the <first massive level of the open RNS string> has
$$
\boxed{\alpha'M^2=1:\qquad8+64+56=128\ \text{bosons}=64+64=128\ \text{fermions}.}
$$
This is the open-string mass normalization; the same chiral oscillator count used in a <closed string> would correspond to $\alpha'M^2=4$.
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