Is RSA key size the size of private key exponent? I have implemented a key pair generation scheme for RSA algorithm. I have taken the length of private key exponent as RSA key size, but then I've got to know that RSA key size is the size of the modulus $n$. Is it wrong to take RSA key size as the size of the private key exponent?

To generate a private key of 1024 bits, I increase the size of private key exponent regardless the modulus $n$.
No, the RSA key size is not the size of the private key exponent. It is customarily the number of bits in the public modulus (which is known as $N$). In other words, the key size is the integer $k$ such that $2^{k-1}\le N&lt;2^k$.

In most implementations (and all implementations conforming to PKCS#1), a private exponent $d$ has size in bits at most the key size $k$, and typically is few bits smaller. Notice that the private exponent is not uniquely defined.

The smallest private exponent is $d=e^{-1}\bmod\lambda(N)$. It is always less than $\lambda(N)$, where $\lambda()$ is Carmichael's function, with $\lambda(N)=\operatorname{lcm}(p-1,q-1)$ when $N=p\cdot q$ with $p$ and $q$ distinct primes. The smallest private exponent always has size in bits strictly less than the key size $k$, and typically is a few bits smaller. 

Sometime, a private exponent is computed as $d=e^{-1}\bmod\varphi(N)$, where $\varphi()$ is Euler's function, with $\varphi(N)=(p-1)\cdot(q-1)$ when $N=p\cdot q$ with $p$ and $q$ distinct primes. That particular $d$ has size in bits at most the key size $k$, often is a few bits smaller, and often is a few bits larger than the smallest private exponent.

Yet other methods to define a private exponent give no maximum for $d$, and only require that $d\equiv e^{-1}\pmod{\lambda(N)}$ (which is the necessary and sufficient condition for $d$ to work), or $d\equiv e^{-1}\pmod{\varphi(N)}$.