Answer: $2^{n-1}$
No. of subsets with cardinality $i = {}^nC_i$
So, no. of subsets with odd cardinality = $\sum_{i = 1, 3, \dots, n-1} {}^nC_i \\ = 2^{n-1} \text{(Proof given below)} $
We have,
${}^nC_0+ {}^nC_1 + {}^nC_2 + \dots + {}^nC_n = 2^n$
${}^nC_0+ {}^nC_1 + {}^nC_2 + \dots + {}^nC_n =\begin{cases} {}^{n+1}C_1+ {}^{n+1}C_3+ \dots +{}^{n+1}C_n, n \text{ is even } \\ {}^{n+1}C_1+ {}^{n+1}C_3+ \dots +{}^{n+1}C_{n-1} +{}^{n}C_{n}, n \text{ is odd } \end{cases}$
$\left(\because {}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r \right) = 2^n$
$\implies \left.\begin{matrix} {}^{n}C_1+ {}^{n}C_3+ \dots +{}^{n}C_{n-1}, n \text{ is even} \\ {}^{n}C_1+ {}^{n}C_3+ \dots +{}^{n}C_{n}, n \text{ is odd} \end{matrix} \right\}= 2^{n-1} \left(\text{ replacing }n \text{ by }n-1, {}^nC_n = {}^{n-1}C_{n-1} \right)$
Proof for ${}^{n}C_r + {}^{n}C_{r-1} = {}^{n+1}C_r$
${}^{n}C_r + {}^{n}C_{r-1} = \frac{n!}{r! (n-r)!} + \frac{n!}{(r-1)! (n-r+1)!}$
$= \frac{ n! (n-r+1) + n! r} {r! (n-r+1)!}$
$= \frac {n! (n+1)}{r! (n-r+1)!}$
$= \frac{(n+1)!}{r! (n-r+1)! } = {}^{n+1}C_r$