Converting a number to 16-bit Floating Point Format
First of all, the IEEE 745 16-bit format has:


1 bit for the sign.
5 bits (not 4) for the biased exponent. The bias is $15$.
10 bits for the mantissa. The leading $1$ is implicit (not explicit).


The calculation is incorrect since $(29.375)_2 = (11101.011)_2 = (1.1101011)_2 \cdot 2^4$ (and not $2^3$).

Then:


The sign bit is $1$ because the number is negative.
The unbiased exponent is $(4)_{10} = (100)_{2}$ and therefore the biased exponent is $(19)_{10} = (10011)_2$.
The mantissa, without the implicit leading $1$ and padded to 10 bits with trailing zeros, is  $(1101011000)_2$.


Then the final $16$-bit number is: $11001101011000$.