I tried doing this with log10() and pow(), but ended up having problems with rounding errors. So as commented by @Karoly Horvath, string manipulation is probably the best approach.

#include <stdlib.h>

char *fortran_sprintf_double(double x, int ndigits) {
  char format[30], *p;
  static char output[30];

  /* Create format string (constrain number of digits to range 1–15) */
  if (ndigits > 15) ndigits = 15;
  if (ndigits < 1) ndigits = 1;
  sprintf(format, "%%#.%dE", ndigits-1);

  /* Convert number to exponential format (multiply by 10) */
  sprintf(output, format, x * 10.0);

  /* Move the decimal point one place to the left (divide by 10) */
  for (p=output+1; *p; p++) {
    if (*p=='.') {
      *p = p[-1];
      p[-1] = '.';
      break;
    }
  }

  return output;
}

If you only care about the integer part being 0 and not really leaving out the 0, i.e. if you're fine with 0.27460E+03 instead of .27460E+03 you could do something similar to this:

#include <stdio.h>
#include <stdlib.h>

void fortran_printf();

int main(void)
{
        double num = 274.600;
        fortran_printf(num);

        exit(EXIT_SUCCESS);
}

void fortran_printf(double num)
{
        int num_e = 0;
        while (num > 1.0) {
                num /= 10;
                num_e++;
        }

        printf("%.5fE+%02d", num, num_e);
}

Otherwise you have to take a detour over strings. Note that the code above is only meant to get you started. It certainly doesn't handle any involved cases.

A string manipulation approach:

int printf_NoIntegerPart(double x, int prec) {
  char buf[20 + prec];
  sprintf(buf, "%+.*E", prec - 1, x * 10.0);  // use + for consistent width output
  if (buf[2] == '.') {
    buf[2] = buf[1];
    buf[1] = '.';
  }
  puts(buf);
}

int main(void) {
  printf_NoIntegerPart(2.74600E+02, 5);  // --> +.27460E+03
}

This will print "INF" for |x| > DBL_MAX/10