Can't open config file: /usr/local/ssl/openssl.cnf on Windows [duplicate]

I have installed OpenSSL 64. I want to use a certificate for my nodejs https server. I ran the following command:

openssl genrsa -out subdomain.domain.com.key 1024

But I have got the error:

WARNING: can't open config file: /usr/local/ssl/openssl.cnf
Loading 'screen' into random state - done
Generating RSA private key, 1024 bit long modulus
.........++++++
.........................................++++++
unable to write 'random state'
e is 65537 (0x10001)

How can I resolve it?
Is this the right command?

The solution is running this command:

set OPENSSL_CONF=C:\OpenSSL-Win32\bin\openssl.cfg   

or

set OPENSSL_CONF=[path-to-OpenSSL-install-dir]\bin\openssl.cfg

in the command prompt before using openssl command.

Let openssl know for sure where to find its .cfg file.

Alternatively you could set the same variable OPENSSL_CONF in the Windows environment variables.

NOTE: This can happen when using the OpenSSL binary distribution from Shining Light Productions (a compiled + installer version of the official OpenSSL that is free to download & use). This distribution is "semi-officially" linked from OpenSSL's site as a "service primarily for operating systems where there are no pre-compiled OpenSSL packages".

I've SSL on Apache2.4.4 and executing this code at first, did the trick:
set OPENSSL_CONF=C:\wamp\bin\apache\Apache2.4.4\conf\openssl.cnf

then execute the rest codes..

/usr/local/ssl/openssl.cnf

A path like this means the program has been compiled with either Cygwin or MSYS. If you must use this openssl then you will need an interpreter that understands those paths, like Bash, which is provided by Cygwin or MSYS.

Another option would be to download or compile a Windows Native version of openssl. Using that the program would instead require a path like

C:\Users\Steven\ssl\openssl.cnf

which would be better suited for the Command Prompt.