1. We need the fractional part of $\left\{\frac{2^{403}}{15}\right\}.$

If we write $2^{403}=15q+r,$ where $0\le r<15$, then $\frac{2^{403}}{15}=q+\frac{r}{15}$ and hence the fractional part is $\left\{\frac{2^{403}}{15}\right\}=\frac{r}{15}.$ So we only need to find $2^{403}\pmod{15}.$

2. Compute powers of $2$ modulo $15$: $2^1\equiv 2 \pmod{15},$ $2^2\equiv 4 \pmod{15},$ $2^3\equiv 8 \pmod{15},$ $2^4\equiv 16\equiv 1 \pmod{15}.$

So the powers repeat with period $4$.

3. Reduce the exponent modulo $4$: $403\equiv 3 \pmod{4}.$ Therefore, $2^{403}\equiv 2^3\equiv 8 \pmod{15}.$ Thus the remainder is $r=8$.

4. Hence, $\left\{\frac{2^{403}}{15}\right\}=\frac{8}{15}.$ Given that this fractional part is $\frac{k}{15}$, we get $k=8.$

5. Checking options:

• A: $8$ ✅
• B: $14$ ❌
• C: $6$ ❌
• D: $1$ ❌

Therefore, the correct answer is A.